lim(sqx^2+2 - sqx^2-6x) for x->infinity

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I think you mean `sqrt(x^2+2)-sqrt(x^2-6x).`

To remove the radicals, let us multiply the top and the bottom of this expression by the sum of the square roots. This will remove the square roots from the numerator.

`sqrt(x^2+2)-sqrt(x^2-6x)=((sqrt(x^2+2)-sqrt(x^2-6x))*(sqrt(x^2+2)+sqrt(x^2-6x)))/(sqrt(x^2+2)+sqrt(x^2-6x)).`

This is equal to

`((x^2+2)-(x^2-6x))/(sqrt(x^2+2)+sqrt(x^2-6x))=(6x+2)/(sqrt(x^2+2)+sqrt(x^2-6x)).`

Now if we multiply the top and the bottom by 1/x we can simplify this into something we can take the limit of.

`(6x+2)/(sqrt(x^2+2)+sqrt(x^2-6x))=(1/x(6x+2))/(1/x (sqrt(x^2+2)+sqrt(x^2-6x)))=(6+2/x)/(sqrt((x^2+2)/x^2)+sqrt((x^2-6x)/x^2))`

It is in turn equal to  `(6+2/x)/(sqrt(1+2/x^2)+sqrt(1-6/x))`

and here all terms have limits:

`(6+0)/(sqrt(1+0)+sqrt(1-0))=3` when `x->+oo,` and `-(6+0)/(sqrt(1+0)+sqrt(1-0))=-3` when `x->-oo.`

 

 

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